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id="post-info"><h1 class="post-title">Gamma 函数详解</h1><div id="post-meta"><div class="meta-firstline"><span class="post-meta-date"><i class="far fa-calendar-alt fa-fw post-meta-icon"></i><span class="post-meta-label">发表于</span><time class="post-meta-date-created" datetime="2020-04-08T11:37:10.000Z" title="发表于 2020-04-08 19:37:10">2020-04-08</time><span class="post-meta-separator">|</span><i class="fas fa-history fa-fw post-meta-icon"></i><span class="post-meta-label">更新于</span><time class="post-meta-date-updated" datetime="2021-08-14T04:39:34.425Z" title="更新于 2021-08-14 12:39:34">2021-08-14</time></span><span class="post-meta-categories"><span class="post-meta-separator">|</span><i class="fas fa-inbox fa-fw post-meta-icon"></i><a class="post-meta-categories" href="/categories/%E6%95%B0%E5%AD%A6/">数学</a><i class="fas fa-angle-right post-meta-separator"></i><i class="fas fa-inbox fa-fw post-meta-icon"></i><a class="post-meta-categories" href="/categories/%E6%95%B0%E5%AD%A6/%E6%A6%82%E7%8E%87%E8%AE%BA/">概率论</a></span></div><div class="meta-secondline"><span class="post-meta-separator">|</span><span class="post-meta-pv-cv"><i class="far fa-eye fa-fw post-meta-icon"></i><span class="post-meta-label">阅读量:</span><span id="busuanzi_value_page_pv"></span></span><span class="post-meta-separator">|</span><span class="post-meta-commentcount"><i class="far fa-comments fa-fw post-meta-icon"></i><span class="post-meta-label">评论数:</span><a href="/math/probability_theory/gamma/#post-comment"><span id="twikoo-count"></span></a></span></div></div></div></header><main class="layout" id="content-inner"><div id="post"><article class="post-content" id="article-container"><h1 id="简介"><a href="#简介" class="headerlink" title="简介"></a>简介</h1><p>&nbsp;$\mathrm{Gamma}$ 函数如下定义<br>$$<br>\Gamma(x) = \int_0^{\infty}t^{x - 1}e^{-t},\mathrm{d}t<br>$$<br>那么它是如何来的呢?</p>
<p>考虑如下的一个数列<br>$$<br>a_n = {1, 4, 9, 16, \cdots}<br>$$<br>这个数列可以用通项公式 $n^2$ 来表达, 即使 $n$ 为实数也是良好定义的. 现在我们来考虑阶乘 $n!$ . 在 $n$ 为整数时, 它是良好定义的, 但是当 $n$ 为实数时呢? $0.5!$ 等于多少? 欧拉于 1729 年完美的解决了这个问题, 由此导致了 $\mathrm{Gamma}$ 函数的产生. $\mathrm{Gamma}$ 函数可以看作是定义在实数域的 $(n - 1)!$ 为什么不是 $n!$ 呢? 后面会说明.</p>
<p>欧拉考虑了如下形式的积分 (至于为什么是<strong>如下形式</strong>的<strong>积分</strong>, 请自行了解).<br>$$<br>J(e,n) = \int_0^1 x^e(1 - x)^n,\mathrm{d}x<br>$$<br><strong>这里的 $e$ 并不是自然对数的底数, 而是任意实数.</strong> 由分布积分可得<br>$$<br>\begin{aligned}<br>\int_0^1x^e(1 - x)^n\mathrm dx &amp;= \int_0^1\left(\frac{1}{e + 1}x^{e+1}\right)’(1 - x)^n,\mathrm{d}x\\<br>&amp;=\left.\left(\left(\frac{1}{e+1}x^{e+1}\right)(1 - x)^n\right)\right|^1_0 - \int_0^1\left(\frac{1}{e+1}x^{e+1}\right)\cdot-n(1-x)^{n - 1},\mathrm{d}x\\<br>&amp;=\frac{n}{e+1}\int_0^1 x^{e + 1}(1 - x)^{n - 1}<br>\end{aligned}<br>$$<br>因此<br>$$<br>J(e, n) = \frac{n}{e+1}J(e+1, n-1)<br>$$<br>不断进行这个过程, 最终会得到<br>$$<br>\begin{aligned}<br>J(e, n) &amp;= \frac{n(n - 1)(n - 2)\cdots1}{(e+1)(e+2)(e + 3)\cdots(e+n)}J(e+n, 0)\\<br>&amp;= \frac{n(n - 1)(n - 2)\cdots1}{(e+1)(e+2)(e + 3)\cdots(e+n)}\left.\left(\frac{1}{e+n+1}x^{e+n+1}\right)\right|^1_0\\<br>&amp;=\frac{n!}{(e+1)(e+2)(e + 3)\cdots(e+n)(e+n+1)}<br>\end{aligned}<br>$$<br>因此有<br>$$<br>n! = (e+1)(e+2)\cdots(e+n+1)J(e,n)\tag{1}<br>$$<br>此时就已经可以表达实数域的 $n!$ 了.</p>
<p>为了式子更加简化, 欧拉运用了<del>亿点点</del>计算技巧, 取 $e = f/g$ 并令 $f\rightarrow1, g\rightarrow0$ .</p>
<p>首先令 $x = t^{\frac{g}{f+g}}$ , 而有<br>$$<br>\begin{aligned}<br>\frac{,\mathrm{d}x}{,\mathrm{d}t} &amp;= \frac{g}{f+g}t^{\frac{g}{f+g} - 1}\\<br>\mathrm dx &amp;= \frac{g}{f+g}t^{-\frac{f}{f+g}}\mathrm dt<br>\end{aligned}<br>$$</p>
<p>当 $x\in[0,1]$ 时, $t\in[0, 1]$ .</p>
<p>带入式 $(1)$ 可得<br>$$<br>\begin{aligned}<br>n! &amp;= (e+1)(e+2)\cdots(e+n+1)\int_0^1 x^e(1 - x)^n,\mathrm{d}x\\<br>&amp;= (\frac{f}{g}+1)(\frac{f}{g}+2)\cdots(\frac{f}{g}+n+1)\int_0^1t^{\frac{f}{f+g}}(1 - t^{\frac{g}{f+g}})^n \frac{g}{f+g}t^{-\frac{f}{f+g}},\mathrm{d}t\\<br>&amp;=(\frac{f}{g}+1)(\frac{f}{g}+2)\cdots(\frac{f}{g}+n+1)\int_0^1(1 - t^{\frac{g}{f+g}})^n\frac{g}{f+g},\mathrm{d}t\\<br>&amp;=\frac{(f+g)(f+2g)\cdots(f+(n+1)g)}{g^{n+1}}\int_0^1\left(\frac{1-t^{\frac{g}{f+g}}}{\frac{g}{f+g}}\right)^n\left(\frac{g}{f+g}\right)^{n + 1},\mathrm{d}t\\<br>&amp;=\frac{(f+g)(f+2g)\cdots(f+(n+1)g)}{(f+g)^{n+1}}\int^1_0\left(\frac{1-t^{\frac{g}{f+g}}}{\frac{g}{f+g}}\right)^n,\mathrm{d}t<br>\end{aligned}<br>$$<br>积分左边显然等于 $1$ , 那么右边等于什么呢?</p>
<p>上式等价于<br>$$<br>\int^1_0\left(\lim_{x\rightarrow0}\frac{1-t^{x}}{x}\right)^n,\mathrm{d}t<br>$$<br>使用洛必达法则<br>$$<br>\begin{aligned}<br>\int^1_0\left(\lim_{x\rightarrow0}\frac{1-t^{x}}{x}\right)^n,\mathrm{d}x &amp;= \int^1_0\left(\lim_{x\rightarrow0}-t^x\ln t\right)^n,\mathrm{d}t\\<br>&amp;= \int_0^1(-\ln t)^n,\mathrm{d}t<br>\end{aligned}<br>$$<br>于是 $n!$ 就简化成了<br>$$<br>n! =\int_0^1(-\ln t)^n,\mathrm{d}t<br>$$<br>令 $t = \mathrm{e}^{-u}$ , <strong>(注意这里的 $e$ 为自然对数的底数)</strong> , 当 $t\in[0, 1]$ 时 $u\in [0, +\infty]$ .</p>
<p>有<br>$$<br>\begin{aligned}<br>\frac{,\mathrm{d}t}{,\mathrm{d}u} &amp;= (\mathrm{e}^{-u})’\\<br>\frac{,\mathrm{d}t}{,\mathrm{d}u}&amp;= -\mathrm{e}^{-u}\\<br>,\mathrm{d}t&amp;=-\mathrm{e}^{-u},\mathrm{d}u<br>\end{aligned}<br>$$<br>这里要注意到 $\mathrm{d}u &lt; 0$ .</p>
<p>因此<br>$$<br>\begin{aligned}<br>n! &amp;= \int_0^{+\infty} u^n\mathrm{e}^{-u},\mathrm{d}u<br>\end{aligned}<br>$$<br>由于积分中的 $\mathrm{d}u&gt;0$ , 因此负号被抵消了. </p>
<p>这就是 $n!$ 的积分表达, 而<br>$$<br>\Gamma(x) = \int_0^{\infty}t^{x - 1}\mathrm{e}^{-t},\mathrm{d}t = (n-1)!<br>$$<br>所以, 为什么不是 $n!$ 呢? 欧拉最早的 $\mathrm{Gamma}$ 函数定义还真是 $\Gamma(n) = n!$ 但是欧拉后来不知道出于什么原因, 修改了 $\mathrm{Gamma}$ 函数的定义, 使得 $\Gamma(n) = (n-1)!$ 有数学家猜测可能是欧拉研究了如下形式的积分<br>$$<br>\begin{aligned}<br>B(m,n)&amp;=\int_0^1x^{m-1}(1-x)^{n-1},\mathrm{d}x\\<br>&amp;=\frac{\Gamma(m)\Gamma(n)}{\Gamma(m+n)}<br>\end{aligned}<br>$$<br>这个函数现在称为 $\mathrm{Beta}$ 函数. 看, 多么有美感的一个式子, 如果 $\Gamma(n) = n!$ , 那么就会变为<br>$$<br>B(m, n) = \frac{\Gamma(m)\Gamma(n)}{\Gamma(m+n+1)}<br>$$<br>公式的美感就降低了.</p>
<p>其证明可以看我的<a href="https://yunist.cn/math/probability_theory/beta/">这篇文章</a>.</p>
<p>不过, 这只是一个定义, 并不需要太纠结.</p>
<hr>
<h1 id="参考资料"><a href="#参考资料" class="headerlink" title="参考资料"></a>参考资料</h1><h2 id="网络资料"><a href="#网络资料" class="headerlink" title="网络资料"></a>网络资料</h2><ul>
<li><p><a target="_blank" rel="noopener" href="https://www.jianshu.com/p/387ab7b9998b">Gamma 函数推导</a></p>
</li>
<li><p><a target="_blank" rel="noopener" href="http://bloglxm.oss-cn-beijing.aliyuncs.com/lda-LDA%E6%95%B0%E5%AD%A6%E5%85%AB%E5%8D%A6.pdf">LDA 数学八卦</a></p>
</li>
</ul>
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